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High order correlations in sampling and concentration bounds via size biasing
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High order correlations in sampling and concentration bounds via size biasing
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Copyright 2022 Christopher Wayne Walker HIGH ORDER CORRELATIONS IN SAMPLING AND CONCENTRATION BOUNDS VIA SIZE BIASING by Christopher Wayne Walker A Dissertation Presented to the FACULTY OF THE USC GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulfillment of the Requirements for the Degree DOCTOR OF PHILOSOPHY (APPLIED MATHEMATICS) August 2022 To my wife Lorri, for her enduring patience and understanding ii Acknowledgments I wouldlike to express my sincere appreciationto my majoradvisor,ProfessorLarry Goldstein, for his time and valuable suggestions throughout this study. His encouragement helped motivate me to complete this work. I would also like to thank Professors Ken Alexander, Keith Chugg, Jason Fulman and Nicolai Haydn for serving on my Qualifying Exam committee with Professors Chugg and Haydn also serving on my Dissertation Committee. Finally, I express my thanks to Professor SusanMontgomery,theMathematicsDepartmentDirectorofGraduateStudies,forallhervaluable assistance. 1 Table of Contents Dedication ii Acknowledgments iii List of Tables v List of Figures vi Abstract vii Chapter 1: High Order Correlations In Sampling 1 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Chapter 2: Concentration Bounds via Size Biasing 26 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 2.2 Prior Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.3 Upper Tail Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.3.1 Bounded case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.3.2 Analytic Comparison of Bounds . . . . . . . . . . . . . . . . . . . . . . . . . 37 2.3.3 Unbounded Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 2.4 Lower Tail Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 2.4.1 Bounded Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 2.4.2 Unbounded Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 2.5 Practical Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 2.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 Bibliography 66 iv List of Tables 2.1 Example for Size Biasing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 v List of Figures 2.1 Ratio of Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 2.2 Zoomed in Ratio of Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 2.3 Coverage Example. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 2.4 Sliding Window Example. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 2.5 Urn Allocation Example.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 2.6 Length m Runs Example. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 2.7 M of N Example. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 vi Abstract For N = 1,2,..., letS N be a simple random sample of size n = n N from a populationA N of size N, where 0≤n≤N. Then with f N =n/N, the sampling fraction, and1 A the inclusion indicator that A∈S N , for any H ⊂A N of size k≥0, the high order correlations Corr(k)=E Y A∈H (1 A −f N ) ! depend only on k, and if the sampling fraction f N →f as N →∞, we show in this thesis that N k/2 Corr(k)→[f(f−1)] k/2 EZ k , k even and N (k+1)/2 Corr(k)→[f(f−1)] (k−1)/2 (2f−1) 1 3 (k−1)EZ k+1 , k odd where Z is a standard normal random variable. Suppose X ≥0 with EX =a>0 and let Y be a size biased version of X such that Y ≤X +c for some c∈ (0,∞). Let k =b |x−a| c c, I a,c = a 2 −ac+c 2 a−c , a6=c. Define for x≥a, U P (x,a,c):= Y 0≤i≤k a x−ic (the upper tail product bound) and define for 0≤x≤a, vii L P (x,a,c):= Y 0a, x−kc≥a U PC,Δ (x,a,c):= ac ac+(a−x+(k−Δ+1)c) 2 Y 0≤i≤k−Δ a x−ic (1) and define for Δ≥0, 0≤xk−m+v, and P 0 j,v (1)=P j,v (1), P 0 k,m−v (j)=P k,m−v (j) for 0≤j≤k. At this point in our proof we will consider k even and k odd separately. For k even it will be convenient to write (1.25) and (1.26), respectively, as P k,m (j)=(−1) m j 2m 2 m m! +r (k,m),2m−1 (j), and P j,v (1)= j 2v 2 v v! +r v,2v−1 (j). 19 Substituting these last two expressions into (1.41) we get α(k,f N ;k−v) = (−1) k k+v X m=v k X j=0 " k j (−1) j+m j 2v 2 v v! +r v,2v−1 (j) × (−1) m−v j 2(m−v) 2 m−v (m−v)! +r (k,m,v),2(m−v)−1 (j) !# N k−m = (−1) k k+v X m=v k X j=0 " k j (−1) j+m × (−1) m−v j 2m 2 m v!(m−v)! +r (k,m,v),2m−1 (j) !# N k−m . (1.42) From (1.39) and (1.42) we obtain N k/2 Corr(k,f N ;k−v) = N k/2 α(k,f N ;k−v) (N) k = N k/2 (N) k (−1) k k+v X m=v k X j=0 " k j (−1) j+m × (−1) m−v j 2m 2 m v!(m−v)! +r (k,m,v),2m−1 (j) !# N k−m = N k/2 (N) k (−1) k−v v! k+v X m=v 1 2 m (m−v)! k X j=0 k j (−1) j j 2m (1.43) + k X j=0 k j (−1) j r (k,m,v),2m−1 (j) N k−m . Using (1.8), (1.43) becomes N k/2 Corr(k,f N ;k−v) = N k/2 (N) k (−1) k−v v! k+v X m=v 1 2 m (m−v)! (−1) k k! 2m k (1.44) + k+v X m=v k X j=0 k j (−1) j r (k,m,v),2m−1 (j) ! N k−m . 20 Since (N) k is of degree k, then when m>k/2, lim N→∞ N k/2 (N) k N k−m = lim N→∞ N k (N) k N k/2 N m =0 and when m≤k/2, from (1.8) and (1.9) we have that k X j=0 k j (−1) j j p =(−1) k k! n p k o =0 for all p≤2m−1. Therefore, by linearity, k X j=0 k j (−1) j r (k,m,v),2m−1 (j)=0 for m≤k/2, and since for k even, 2m k =0 for m<k/2, then upon letting N →∞, (1.44) is zero for v >k/2 and for v≤k/2 (1.44) becomes lim N→∞ N k/2 Corr(k,f N ;k−v) (1.45) = lim N→∞ N k/2 (N) k (−1) k−v v! k/2 X m=k/2 1 2 m (m−v)! (−1) k k! 2m k N k−m . When m=k/2, lim N→∞ N k/2 (N) k N k−m = lim N→∞ N k (N) k N k/2 N m = lim N→∞ N k (N) k N k/2 N k/2 =1. Therefore, using (1.9), (1.45) becomes lim N→∞ N k/2 ·Corr(k,f N ;k−v) = (−1) v k! v!2 k/2 k 2 −v ! =(−1) v k! 2 k/2 k 2 ! k/2 v =(−1) v EZ k k/2 v . 21 Now, by our previous notation, Corr(k)=Corr(k,f N ), so lim N→∞ N k/2 ·Corr(k) = lim N→∞ N k/2 ·Corr(k,f N )= lim N→∞ N k/2 k X v=0 Corr(k,f N ;k−v)f k−v N = k/2 X v=0 k/2 v f k−v (−1) v EZ k =[f(f−1)] k/2 EZ k which proves the theorem for k even. For k odd we have from (1.25), (1.26) and (1.41) that α(k,f N ;k−v) = (−1) k k+v X m=v k X j=0 k j (−1) j+m j 2v − 1 3 v(2v+1)j 2v−1 2 v v! +r v,2v−2 (j) ! × " (−1) m−v j 2(m−v) + 1 3 (m−v)(2(m−v)−5)j 2(m−v)−1 2 m−v (m−v)! ! +r (k,m,v),2(m−v)−2 (j) # N k−m which becomes α(k,f N ;k−v) = (−1) k k+v X m=v k X j=0 k j (−1) j+m × " (−1) m−v j 2m + 1 3 [(m−v)(2m−2v−5)−v(2v+1)]j 2m−1 2 m v!(m−v)! ! +r (k,m,v),2m−2 (j) # N k−m . 22 From (1.39) and (1.42) we obtain N (k+1)/2 Corr(k,f N ;k−v)=N (k+1)/2 α(k,f N ;k−v) (N) k = N (k+1)/2 (N) k (−1) k k+v X m=v k X j=0 k j (−1) j+m × " (−1) m−v j 2m + 1 3 [(m−v)(2m−2v−5)−v(2v+1)]j 2m−1 2 m v!(m−v)! ! +r (k,m,v),2m−2 (j) # N k−m = N (k+1)/2 (N) k (−1) k−v v! k+v X m=v 1 2 m (m−v)! k X j=0 k j (−1) j (1.46) × j 2m + 1 3 [(m−v)(2m−2v−5)−v(2v+1)]j 2m−1 + k X j=0 k j (−1) j r (k,m,v),2m−2 (j) N k−m . Using (1.8), (1.46) becomes N (k+1)/2 Corr(k,f N ;k−v) = N (k+1)/2 (N) k (−1) k−v v! " k+v X m=v 1 2 m (m−v)! (−1) k k! 2m k (1.47) + k+v X m=v 1 3 [(m−v)(2m−2v−5)−v(2v+1)] 2 m (m−v)! (−1) k k! 2m−1 k + k+v X m=v k X j=0 k j (−1) j r (k,m,v),2m−2 (j) N k−m . Since (N) k is of degree k, then when m>(k+1)/2, lim N→∞ N (k+1)/2 (N) k N k−m = lim N→∞ N k (N) k N (k+1)/2 N m =0 and when m≤(k+1)/2, from (1.8) and (1.9) we have that k X j=0 k j (−1) j j p =(−1) k k! n p k o =0 for all p≤2m−2. 23 Therefore, by linearity, k X j=0 k j (−1) j r (k,m,v),2m−2 (j)=0 for m≤(k+1)/2, and since for k odd, 2m k = 0 for m<(k+1)/2 and 2m−1 k = 0 for m<(k+1)/2, then upon letting N →∞ (1.47) is zero for v >(k+1)/2 and for v≤(k+1)/2 (1.47) becomes lim N→∞ N (k+1)/2 Corr(k,f N ;k−v) = lim N→∞ N (k+1)/2 (N) k (−1) k−v v! (k+1)/2 X m=(k+1)/2 1 2 m (m−v)! (−1) k k! 2m k (1.48) + (k+1)/2 X m=(k+1)/2 1 3 [(m−v)(2m−2v−5)−v(2v+1)] 2 m (m−v)! (−1) k k! 2m−1 k N k−m . When m=(k+1)/2, lim N→∞ N (k+1)/2 (N) k N k−m = lim N→∞ N k (N) k N (k+1)/2 N m = lim N→∞ N k (N) k N (k+1)/2 N (k+1)/2 =1. Therefore,(1.48) becomes lim N→∞ N (k+1)/2 ·Corr(k,f N ;k−v) = (−1) k−v v! " 1 2 (k+1)/2 k+1 2 −v ! (−1) k k! k+1 k + 1 3 k+1 2 −v (k+1−2v−5)−v(2v+1) 2 (k+1)/2 k+1 2 −v ! (−1) k k! k k # . Using (1.9) this last result becomes lim N→∞ N (k+1)/2 ·Corr(k,f N ;k−v) = 1 3 (−1) v k! v!2 (k+1)/2 k+1 2 −v ! 3 k+1 2 + k+1 2 −v (k−2v−4)−v(2v+1) = 2 3 (−1) v (k−1)(k+1−v)k! v!2 (k+1)/2 k+1 2 −v ! . 24 We may write this last result as lim N→∞ N (k+1)/2 ·Corr(k,f N ;k−v) = 2 3 (−1) v (k−1)(k+1−v)k! k+1 2 v 2 (k+1)/2 k+1 2 ! = 2 3 (−1) v (k−1)(k+1−v)EZ k+1 k+1 k+1 2 v . Hence, lim N→∞ N (k+1)/2 ·Corr(k) = lim N→∞ N (k+1)/2 ·Corr(k,f N ) = lim N→∞ N (k+1)/2 k X v=0 Corr(k,f N ;k−v)f k−v N = k X v=0 2 3 (−1) v (k−1)(k+1−v) k+1 k+1 2 v f k−v EZ k+1 = 2 3 (k−1) k+1 EZ k+1 d df (k+1)/2 X v=0 (−1) v k+1 2 v f k+1−v = 2 3 (k−1) k+1 EZ k+1 d df [f(f−1)] (k+1)/2 = [f(f−1)] (k−1)/2 (2f−1) 1 3 (k−1)EZ k+1 which proves the theorem for k odd. The proof of the theorem is now complete. 25 Chapter 2 Concentration Bounds via Size Biasing 2.1 Introduction Concentration bounds (or inequalities) refer to the bounding of probabilities (either the upper tail or the lower tail). In some circumstances one can compute for a random variable X and a real number x 0 , P(X > x 0 ) or P(X < x 0 ), exactly in a closed form expression by knowing the underlying distribution function for X. One may need to define what constitutes a closed form expression. In the best case this would involve an algebraic expression only, that is, one using a finite number ofthe usual algebraicoperations. The next best result we mighthope for is a defined but not an algebraicexpression such as expressionsinvolvingthe Gaussiantailfunction,Q(x). But for the task at hand we might not know the underlying distribution or, even if it is known, the cumulative distribution function might be too complicated to easily use in analysis. Concentration bounds provide a technique not to compute probabilities exactly but rather to bound themin such a waythatthe resultingbound stillsufficientlyaidsus in analysis. Alongthese lines we may have inequalities that apply under certain conditions. The more general the condi- tions the more applicable the inequality could potentially be to a host of problems. In particular, concentration inequalities based on size bias couplings are fairly general. In this chapter we investigate some concentration bounds derived using the concept of size biasing. In size biasing random variables we sample them proportional to their size. This is illustrated with the following example taken from [3]. Example: Students who ate at the school cafeteria are asked ”How many people, including yourself, sat at your table?” The results were 20% said they ate alone, 30% said they ate with 1 other person, 30% said they ate at a table of 3 and 20% said they ate at a table of 4. From this information we cannot conclude that 20% of the tables had 1 person, 30% had 2 people, 30% had 3 people and 20% had 4 people. Imagine 100 students were in this sample. Then, 26 20 students ate alone using 20 tables, 30 students ate in pairs using 15 tables, 30 students ate in groups of 3 using 10 tables, and 20 students ate in groups of 4 using 5 tables. Thus, there were 20+15+10+5=50occupied tablesofwhich 40%had1 person, 30%had2 people, 20%had 3people, and 10% had 4 people. Let us now consider an experiment where we select an occupied table at random and let the random variable X denote the number of students at the selected table. Let us also consider an experiment where we select a student at random and let X ∗ denote the total number of students at the table where the selected student sat. Based on the previous information we can construct Table 2-1 on the following page. We see that for X each table has the same chance of being selected and for X ∗ the chance to select a table is proportional to the number of people who sat there so the distributions for X and X ∗ are related. We conclude that P(X ∗ =k) is proportional to k×P(X =k). Hence, we see that P(X ∗ =k)=c×k×P(X =k),forsomec. Using1 = P k P(X ∗ =k) =c P k kP(X =k)=c×EX we get c=1/EX. Thus, P(X ∗ =k) = kP(X =k) EX , k =0,1,2,... From this last expression we see that the distribution of X ∗ is determined by the weight or size of X and say that X ∗ has the X size biased distribution. The example above illustratesthe concept of size biasing. We used random variables that took on integer values but this concept also applies in the continuous case. More generally, from [3] a random variableX can be size biased ifand onlyif itis nonnegative,with finiteand positivemean. We will let a =EX,a∈ (0,∞). We say X ∗ has the X size biased distribution if and only if for all bounded continuous functions g, Eg(X ∗ )= 1 a E(Xg(X)). Specifically, in workingwiththe product bounds from[2]weareconcerned atfirstwithcases where X ∗ ≤X+c for some nonnegative constant c. Then later we will relax this requirement somewhat and consider X ∗ ≤X +c (with some probability). 27 . k P(X =k) P(X ∗ =k) 1 2 3 4 0.4 0.3 0.2 0.1 0.2 0.3 0.3 0.2 1.0 1.0 Table 2.1: Example for Size Biasing. The product bounds in [2] are very strong, in fact, they are sharp for the Poisson distribution undercertainconditionssotheycannotbeimprovedinanysignificantwayasfarastheirasymptotic behavior is concerned and still apply in a fairly general setting. However, there is room to improve constantmultiplierstomaketheboundsacertainpercentagetighterandthatiswhatweaccomplish in this research. There is a small cost to pay for this improvement in that a less general restriction is imposed on the mean. New concentration bound results are provided for the case X ∗ ≤ X +c and also for the case X ∗ ≤X +c with some probability. A treatment of concentration inequalities that are not derived via size biasing can be found in [6]. A theoretical treatment of large deviations is found in [9], [16] and [17]. A monograph of concentration inequalities that includes several applications is [18]. The utilization of Stein’s method in concentration inequalities is treated in [7]. However, our emphasis in this study is on probability bounds derived via size biasing random variables. 2.2 Prior Results In this section we first provide some definitions and results from [2] and [8] for convenience of the reader. For any nonnegative random variable X with 0 < EX < ∞ the distribution of Y is said to have the size biased distribution of X, written Y = d X ∗ , if the Radon-Nikodym derivative of the distributionofY,withrespecttothedistributionofX,isgivenbyP(Y ∈dx)/P(X∈dx)=x/EX. If Y = d X ∗ , then for all measurable g, Eg(Y) =E(Xg(X))/EX. X is said to admit a c-bounded size bias coupling (BSBC) if X≥0, EX =a and forY = d X ∗ there exists a coupling in which Y ≤X +c for some c∈ (0, ∞). 28 We now present the product bound development from [2] for the upper tail. Assume BSBC. Then Y ≥x is a subset of the event X ≥x−c. So for x>0 and G(x):=P(X ≥x) xG(x) = xE1(X≥x) ≤ E(X1(X≥x)) = aP(Y ≥x) ≤ aG(x−c) and thus G(x)≤ a x G(x−c). (2.1) Given x>0, one can now iterate (2.1) as long as x 0 =x−ic, i∈Z, i≥0, gives a ratio a/x 0 ≤ 1. This leads to the strong concentration upper tail bound given in [2] repeated here: G(x)≤ Y 0≤i≤k a x−ic , x≥a, (2.2) where k =b |x−a| c c. We will refer to the bound in (2.2) as theupper tail product bound in this study. Suppose EX = a and Var(X)= σ 2 . Then, EX 2 = aEY and Y ≤ X +c implies EY ≤ a+c. Thus, EX 2 =aEY ≤a(a+c) and then σ 2 ≤ac. We will combine this with the one-sided Chebyshev bound (upper tail) ∀x≥a, G(x)≤ σ 2 σ 2 +(a−x) 2 . (2.3) Since σ 2 ≤ac we can now write ∀x≥a, G(x)≤ ac ac+(a−x) 2 . (2.4) Even though this latter form for Chebyshev is looser than the former there are some mathematical advantages in using the latter as regards proving new concentrations bounds as we will see. 29 AgainassumeBSBC. ThenY ≤x isasuperset ofthe eventX ≤x−c. So withF(x):=P(X ≤ x) xF(x) = xE1(X≤x) ≥ E(X1(X≤x)) = aP(Y ≤x) ≥ aF(x−c) and thus F(x)≤ x+c a F(x+c). (2.5) Givenx>0,onecannowiterate(2.5)aslongasx 0 =x−ic, i∈Z, i≥0,givesaratio(x 0 +c)/a≤1. This leads to the strong concentration lower tail bound given in [2] repeated here: F(x)≤ Y 00 then we continue and get pxG(x) = paP(Y ≥x) ≤ aP(Y ≤X +c|Y ≥x)P(Y ≥x) = aP(Y ≤X +c and Y ≥x) ≤ aP(X ≥x−c) ≤ aG(x−c). Thus, pxG(x)≤aG(x−c) which leads to G(x)≤ a/p x G(x−c) (2.10) 31 which is (18) in [8]. This means that in the results above involving the product bound only we can replace a with a/p as indicated in [8] to obtain new concentration inequalities that hold for the unbounded size biased coupling. We say the size biased coupling (X,Y) is c−bounded with probability p for the lower tail if for any x, P(Y ≤X +c|X≤x)≥p. (2.11) The above definition is (11) in [8]. Proceeding as before but incorporating (2.11) we find x p F(x) = x p E1(X≤x) ≥ 1 p E(X1(X≤x)) = a p P(Y ≤x) ≥ a p P(X ≤x−c) ≥ aP(X ≤x−c) = aF(x−c) so F(x)≤ x+c pa F(x+c) (2.12) which is (19) in [8]. This means that in the results above involving the product bound only we can replace a with pa as indicated in [8] to obtain new concentration inequalities that hold for the unbounded size biased coupling. We now present the original result from [11] that utilized size bias couplings to derive concentration inequalities. This theorem includes both the upper tail and lower tail cases. Theorem 2.2.1. Let Y be a nonnegative random variable with mean μ and variance σ 2 , both finite and positive. Suppose there exists a coupling of Y to a variable Y s having the Y−size bias 32 distribution which satisfies|Y s −Y|≤C for some C >0 with probability one. If Y s ≥Y with probability one, then P Y −μ σ ≤−t ≤exp − t 2 2A for all t>0, where A =Cμ/σ 2 . If the moment generating function m(θ)=E[e θY ] is finite at θ =2/C, then P Y −μ σ ≥t ≤exp − t 2 2(A+Bt) for all t>0, (2.13) where A =Cμ/σ 2 and B =C/2σ. Proof: See [11]. 2.3 Upper Tail Bounds In this section we will consider both the bounded and unbounded cases for the upper tail bounds and show how modest improvements can be obtained for the constant multipliers. 2.3.1 Bounded case Theorem 2.3.1. Suppose X ≥ 0 with EX = a > 0 and let Y be a size biased version of X such that Y ≤X +c for some c∈ (0,∞). Let k =b |x−a| c c, I a,c = a 2 −ac+c 2 a−c , a6=c. Define for x≥a, U P (x,a,c):= Y 0≤i≤k a x−ic (the upper tail product bound), and define for Δ≥1, x>a, x−kc≥a U PC,Δ (x,a,c):= ac ac+(a−x+(k−Δ+1)c) 2 Y 0≤i≤k−Δ a x−ic . (2.14) 33 Let G(x):=P(X ≥x). Then, (i) for x≥a, G(x)≤U P (x,a,c), (ii) and for Δ=2, a>2c and x−kc∈ (I a,c , a+c) G(x)≤U PC,2 (x,a,c)(2+ √ 2)c and x−kc∈ (a, I a,c ] G(x)≤U PC,3 (x,a,c)a provided the conditions on a stated in the theorem are met. Proof: Since U P (x,a,c)=U P (x/c,a/c,c/c)and U PC,Δ (x,a,c)=U PC,Δ (x/c,a/c,c/c) we may assume without loss of generality that c=1 in our proof. Part (i) of Theorem 2.3.1 is proven in Arratia and Baxendale [2] and repeated above. For part (ii) we let Δ=2 in (2.14) to get U PC,2 (x,a,1)= a a+(a−x+k−1) 2 Y 0≤i≤k−2 a x−i . 34 By using (2.4) with iteration of (2.1) we get for x>a, k≥1, x−k≥a, G(x)≤ a a+(a−x+k−1) 2 Y 0≤i≤k−2 a x−i (2.15) where we define the empty product to be unity. Now the bound in (2.15) will be better than the upper tail product bound in (2.2) if a a+(a−x+k−1) 2 < Y k−1≤i≤k a x−i . Letting u = x−k analysis of this last expression leads to the following inequality that must be satisfied to prove the claim in the theorem: P 2;a (u)<0, (2.16) where P 2;a (u)=(1−a)u 2 +(2a 2 −2a+1)u−a 3 +a 2 −a. Now P 2;a (u) has roots u 2;1 =a and u 2;2 = a 2 −a+1 a−1 , for a6=1, and we have the single root u 2;1 =1, for a=1. For the latter case (a = 1) the inequality in (2.16) requires u < 1 = a but we require u > a (and a>2) in the theorem so we can dispense with this case. Note that u 2;2 > u 2;1 when a 6= 1 and u 2;2 = I a,c (with c = 1) which is the left end point of interest. We require u > a and a simple derivative test shows that P 2;a (u) is increasing at u = u 2;1 = a. Hence, for this Δ = 2 case the bound will be favorable for u > u 2;2 provided u 2;2 < a+1. It is easy to see that this is indeed the case if a > 2. Injecting arbitrary c back into 35 the analysis we have thus proved the 2nd part of the theorem. For part (iii) we let Δ=3 in (2.14) to get U PC,3 (x,a,1)= a a+(a−x+k−2) 2 Y 0≤i≤k−3 a x−i . By using (2.4) with iteration of (2.1) we get for x>a, k≥2, x−k≥a. G(x)≤ a a+(a−x+k−1) 2 Y 0≤i≤j−2 a x−i . (2.17) Now the bound in (2.17) will be better than the product bound in (2.2) if a a+(a−x+k−2) 2 < Y k−2≤i≤k a x−i . Letting u = x−k analysis of this last expression leads to the following inequality that must be satisfied to prove the claim in the theorem: P 3;a (u)<0, (2.18) where P 3;a (u)=u 3 +(3−a 2 )u 2 +(2−4a 2 +2a 3 )u−4a 2 +3a 3 −a 4 . We find at the left endpoint of interest that P 3;a (a)=a(2−a)<0, provided a>2. We find at the right endpoint of interest that P 3;a a 2 −a+1 a−1 =− a 3 (a 2 −4a+2) (a−1) 3 <0, provided a>2+ √ 2. 36 For a>2+ √ 2 the cubic P 3;a (u) has only one real root given by u 3;1 =A+ B 3(C+D) 1/3 + 1 3×2 1/3 (C+D) 1/3 where A = 1 3 (−3+a 2 ) B = −2 1/3 (−3−6a 2 +6a 3 −a 4 ) C = 36a 2 −27a 3 +45a 4 −18a 5 +2a 6 D = 3 √ 3 p −4−24a 2 +24a 3 −4a 4 +24a 5 +51a 6 −26a 7 +3a 8 (2.19) Since P 3;a (u)has only one realroot (and itisof multiplicity1)and fora>2+ √ 2 bothP 3;a (a)<0 and P 3;a a 2 −a+1 a−1 < 0, we conclude that P 3;a (a) < 0 over the interval of interest and thus (2.18) is satisfied and this proves the third part of the theorem. 2.3.2 Analytic Comparison of Bounds In this section the performance of the bounds is investigated. First we consider the interval given in part (ii) of the theorem corresponding to Δ =2. As before we let c =1 in our analysis. Computingthe ratio,r 2 (u,a),ofthe new product-Chebyshev bound with Δ=2 to the product bound we find for u=x−k r 2 (u,a) = u 2 +u au 2 +(2a−2a 2 )u+a−a 2 +a 3 . Setting ∂r 2 (u,a)/∂u=0 gives two solutions. Choosing the positive solution,u 1 , we find u 1 = 1−a+a 2 + √ a+a 4 −1+2a . It is easy to show with a > 0 that u 1 > a if and only if 2a 2 +a−1 > 0 and since a > 2 we see this condition is alwayssatisfied so that indeed u 1 >a. On the other hand we can show witha>0 that u 1 2. Hence, u 1 always 37 ! " # $ # % $ Figure 2.1: Ratio of Bounds lies in the interval (a, a+1). Evaluating r 2 (u,a) at u =a and u =a+1 gives r 2 (a,a)=1, r 2 (a+1,a)= a 2 +3a+2 a 2 +4a . Note that r 2 (a+1,a)<1 for a>2. Also, at u =I a,c we know r 2 (I a,c ,a)=1. Hence, r 2 (u 1 ,a) is a maximum and r 2 (u,a) is decreasing as u approaches a+1. Now by direct calculationwe see that r 2 (a+1,a) is minimized at a =2+2 √ 3 and the ratio at this point is r 2 (2+2 √ 3+1,2+2 √ 3) = 2+ √ 3 4 ≈0.93. See the two figures for performance as a function of the mean, a. 38 & ’ ( ) * + , - * + , . * + , / * + , 0 * + , 1 * + , 2 * + , , 3 + * * 3 + * 3 & ) 4 5 6 7 & 7 ) 8 9 : ; < = > ? @ A B C A D E E F G H E I J K A L M N ? O P @ N ? Q A E G H E I J K A > E J R I Q @ S T ? C R Figure 2.2: Zoomed in Ratio of Bounds Computing the ratio, r 3 (u,a), of the new bound with Δ =3 to the product bound we find for u =x−k r 3 (u,a) = u 3 +3u 2 +2u a 2 u 2 +(4a 2 −2a 3 )u+4a 2 −3a 3 +a 4 . Setting ∂r 3 (u,a)/∂u= 0 gives four solutions. However, the locations of these solutions do not enable us to ascertain the minimal value of the ratio in the interval of interest. However, we can evaluate the ratio at the endpoints of interest to bound the relative performance there. Firstly, at the left endpoint we have r 3 (a,a) = a 2 +3a+2 a 2 +4a . 39 A direct analysis shows that this ratio is minimized at a =2+2 √ 3 and as above r 3 (2+2 √ 3,2+2 √ 3)= 2+ √ 3 4 ≈0.93. Secondly, at the right endpoint we have r 3 (a+1,a) = a 3 +6a 2 +11a+6 a 3 +9a 2 . An analysis shows that this ratio is minimized at a=11.1285 and we find r 3 (a+1,a)≈0.95. 40 Comparing Δ=2 to Δ =3 It is of interest to know which of the new bounds is best to use in a given situation. Again, specializing c =1 we see from above that Δ=2 will be at least as good as Δ =3 if and only if a a+(a−x+k−1) 2 Y 0≤i≤k−2 a x−i ≤ a a+(a−x+k−2) 2 Y 0≤i≤k−3 a x−i which leads to requiring P(u) = a+(a−u−2) 2 a+(a−u−1) 2 · a u+2 −1≤0 where u =x−k. Now P(u) has 3 roots given by u 1 =a−1, u 2 = 2a−3− √ 4a+1 2 , u 3 = 2a−3+ √ 4a+1 2 , but the only root of interest to us is one that resides in the interval [a, a+1). Thus, the only valid root of P(u) is u 3 . In this case we require a≤u 3 0 and let Y be a c−bounded size biased version of X with probability p such that P(Y ≤ X +c|Y ≥ x)≥ p for some c∈ (0,∞). Further, assume that σ 2 ≤ac. Let k p be defined by k p =b |x−a/p| c c and let I a,c;p = −2ac+2a 2 +c 2 p 2 + p −4a 3 c(1−p 2 )+c 4 p 4 2(a−p 2 c) and let DiscI a,c;p = p −4a 3 c(1−p 2 )+c 4 p 4 denote the discriminant in the expression for I a,c;p . Define for x≥a/p, U P,p (x,a,c):= Y 0≤i≤kp a/p x−ic (the upper tail product bound for unbounded case), and define for Δ≥1, x>a/p, k p ≥1, x−k p c≥a/p U PC,Δ,p (x,a,c):= ac ac+(a−x+(k p −Δ+1)c) 2 Y 0≤i≤kp−Δ a/p x−ic . (2.20) Let G(x):=P(X ≥x) and DiscI a,c;p ≥0. Then, for 42 (i) for x≥a/p, G(x)≤U P,p (x,a,c), (ii) and for Δ=2, k p ≥1, a/p>2c and x−k p c∈ (I a,c;p , a/p+c) G(x)≤U PC,2,p (x,a,c)2+ √ 2 and x−k p c∈ (a/p, I a,c;p ] G(x)≤U PC,3,p (x,a,c)a/p, k≥1, x−k≥a/p, G(x)≤ a a+(a−x+k−1) 2 Y 0≤i≤k−2 a/p x−i (2.21) 43 where we define the empty product to be unity. Now the bound in (2.21) will be better than the product bound for the unbounded case if a a+(a−x+k p −1) 2 < Y kp−1≤i≤kp a/p x−i . (2.22) Letting u = x−k p analysis of this last expression leads to the following inequality that must be satisfied to prove the claim in the theorem: P 2;a,p (u)<0, (2.23) where P 2;a,p (u)=(p 2 −a)u 2 +(2a 2 −2a+p 2 )u−a 3 +a 2 −a. Now P 2;a,p (u) has roots u 2;1 = −2a+2a 2 +p 2 − p −4a 3 +4a 3 p 2 +p 4 2(a−p 2 ) u 2;2 = −2a+2a 2 +p 2 + p −4a 3 +4a 3 p 2 +p 4 2(a−p 2 ) for a 6= p (which is always the case since we require a/p > 2). Note that u 2;2 > u 2;1 when a6= p which always holds for us. In order for the discriminant that appears in the expression for u 2;2 to be nonnegative (to produce a real root) we require p≥ v u u t 2a 3 r 1+ 1 a 3 −1 ! or, equivalently, a≤ 3 s 1 4 · p 4 1−p 2 . (2.24) 44 This lastinequalityputs severe restrictionson a unlessp isvery close to 1. For example, toachieve avalue ofa=2 weneed ptobe about0.986orgreater. On the otherhand, ifa does notsatisfythe inequality in (2.24) then (2.23) will always hold and the product-Chebyshev bound will be better than the product bound for our unbounded size biased coupling. So we now turn our attention to the case where (2.24) is satisfied. We now claim that for a and p satisfying (2.24) with a≥2 and p∈ (0, 1] a/p≤u 2;1 2 which is our area of interest. This provesClaim 1. Claim 2: u 2;2 p p −4a 3 +4a 3 p 2 +p 4 . Again, since we are dealing with real roots the RHS above is positive for a and p satisfying (2.24) and numerical analysis shows that under these same conditions the LHS above is also positive and we can deduce that our claim holds if and only if P 2;2,p (a)<0 where P 2;p,2 (a) = (−4+8p−4p 2 )a 4 +(−16p+20p 2 −8p 3 +4p 4 )a 3 +(−16p 2 +28p 3 −16p 4 )a 2 +(24p 4 −12p 5 )a−8p 6 is a quartic in a. The coefficient of a 4 above is always less than zero if 0 < p < 1 and if p = 1 we are back in the bounded case. So the behavior of this polynomial that we seek to study can be deduced by considering its largest real root. Now P 2;p,2 (a) has 2 real and 2 complex roots. The real roots are a 1 =p 2 and a 2 =c 1 + c 2 c 3 c 4 +3 √ 3 √ c 5 +c 6 1/3 +c 7 c 8 +3 √ 3 √ c 9 +c 10 1/3 46 where c 1 = − 4(−p+p 2 ) 3(−1+2p−p 2 ) c 2 = −(−4p 2 −p 3 +14p 4 −9p 5 ) c 3 = 3(−1+2p−p 2 ) c 4 = −8p 3 +51p 4 −78p 5 +8p 6 +54p 7 −27p 8 c 5 = −32p 7 +167p 8 −303p 9 +130p 10 +263p 11 c 6 = −318p 12 −17p 13 +218p 14 −135p 15 +27p 16 c 7 = 1 3(−1+2p−p 2 ) c 8 = −8p 3 +51p 4 −78p 5 +8p 6 +54p 7 −27p 8 c 9 = −32p 7 +167p 8 −303p 9 +130p 10 +263p 11 c 10 = −318p 12 −17p 13 +218p 14 −135p 15 +27p 16 A numerical analysis of a 2 shows it is negative for all p∈ (0,1) and thus a 1 is the maximum real root and it is less than the permitted minimum a value and thus we conclude that P 2;p,2 (a) < 0 over our region of interest and thus Claim 2 is proved. Therefore, we see that P 2;a,p (u) in (2.23) is indeed negative for u in the interval (u 2;2 , a/p+1) and inserting c back into the analysis proves part (ii) of the theorem. For part (iii) we let Δ=3 in (2.20) to get U PC,3,p (x,a,1)= a a+(a−x+k p −2) 2 Y 0≤i≤kp−3 a/p x−i . By using (2.4) with iteration of (2.10) we get for x>a/p, k≥1, x−k≥a/p, G(x)≤ a a+(a−x+k p −2) 2 Y 0≤i≤kp−3 a/p x−i (2.25) 47 where we define the empty product to be unity. Now the bound in (2.25) will be better than the product bound for the unbounded case if a a+(a−x+k p −2) 2 < Y kp−2≤i≤kp a/p x−i . (2.26) Letting u = x−k p analysis of this last expression leads to the following inequality that must be satisfied to prove the claim in the theorem: P 3;a,p [u]<0, (2.27) where P 3;a,p (u)=p 3 u 3 +(3p 3 −a 2 )u 2 +(2p 3 +2a 3 −4a 2 )u−a 4 +3a 3 −4a 2 . This polynomial has one real root given by u 3;1 =A+ B C(D+ √ E+F) 1/3 +G(D+ √ E+F) 1/3 A = − a 2 +3p 3 3p 3 B = − 2 1/3 (−a 4 −6a 2 p 3 +6a 3 p 3 −3p 6 ) C = 3p 3 D = 2a 6 +18a 4 p 3 −18a 5 p 3 +36a 2 p 6 −27a 3 p 6 +27a 4 p 6 E = 4(−a 4 −6a 2 p 3 +6a 3 p 3 −3p 6 ) 3 F = (2a 6 +18a 4 p 3 −18a 5 p 3 +36a 2 p 6 −27a 3 p 6 +27a 4 p 6 ) 2 G = 1 3×2 1/3 p 3 . NowP 3;a,p (I a,1;p )=0fora =2+ √ 2andp=1. Anumericalinvestigationrevealsthatforthisvalue of a, P 3;a,p (a) < 0 for 0 < p≤ 1 and P 3;a,p (I a,1;p ) < 0 for p∈ [0.997,1). In the case p∈ (0,0.996] 48 we find that I a,1;p is not real and in this case P 3;a,p (u) < 0 for u ∈ [a/p,u 3;1 ) where u 3;1 denotes the real root of P 3;a,p (u) as given above. This proves the third part of the theorem. We now lift the restriction that σ 2 ≤ac in the following. We then have I a,c;p = 2a 3 −2a 2 c+cp 2 σ 2 +σ p −4a 4 +4a 4 p 2 −4a 3 cp 2 +4a 2 p 2 σ 2 +c 2 p 4 σ 2 2(a 2 −p 2 σ 2 ) ·c and we define for Δ≥1, x>a/p, k p ≥1, x−k p c≥a/p U PC,Δ,p (x,a,c):= σ 2 σ 2 +(a−x+(k p −Δ+1)c) 2 Y 0≤i≤kp−Δ a/p x−ic . (2.28) Letting Δ =2 in (2.28) we get U PC,2,p (x,a,1)= σ 2 σ 2 +(a−x+k p c−c) 2 Y 0≤i≤kp−2 a/p x−ic . By using (2.3) with iteration of (2.10) we get for x>a/p, k≥1, x−k≥a/p, G(x)≤ σ 2 σ 2 +(a−x+k p c−c) 2 Y 0≤i≤kp−2 a/p x−ic . Now the new bound will be better than the product bound for the unbounded case if σ 2 σ 2 +(a−x+k p c−c) 2 < Y kp−1≤i≤kp a/p x−ic . (2.29) Letting u = x−k p c analysis of this last expression leads to the following inequality that must be satisfied to prove the claim in the theorem: P 2;a,p,c,σ (u)<0, (2.30) where P 2;a,p,c,σ (u)=(p 2 σ 2 −a 2 )u 2 +(cp 2 σ 2 +2a 3 −2a 2 c)u−a 2 σ 2 −a 4 +2a 3 c−a 2 c 2 . 49 Now P 2;a,p,c,σ (u) has roots u 2;1 = 2a 3 −2a 2 c+cp 2 σ 2 −σ √ D 2(a 2 −p 2 σ 2 ) u 2;2 = 2a 3 −2a 2 c+cp 2 σ 2 +σ √ D 2(a 2 −p 2 σ 2 ) where D =−4a 4 +4a 4 p 2 −4a 3 cp 2 +4a 2 p 2 σ 2 +c 2 p 4 σ 2 and a6=pσ. Letting Δ =3 in (2.28) we get U PC,3,p (x,a,1)= σ 2 σ 2 +(a−x+k p c−c) 2 Y 0≤i≤kp−3 a/p x−ic . Now the new bound will be better than the product bound for the unbounded case if σ 2 σ 2 +(a−x+k p c−c) 2 < Y kp−2≤i≤kp a/p x−ic . (2.31) Letting u = x−k p c analysis of this last expression leads to the following inequality that must be satisfied to prove the claim in the theorem: P 3;a,p,c,σ (u)<0, (2.32) where P 3;a,p,c,σ (u)=p 3 σ 2 u 3 +(−a 3 +3cp 3 σ 2 )u 2 +(−4a 3 +2a 4 +2c 2 p 3 σ 2 )u−4a 3 +4a 4 −a 5 −a 3 σ 2 . The optimal approach then is to take P min (u):=min{P 2;a,p,c,σ (u),P 3;a,p,c,σ (u)} for each u. 50 Forthe bounded case our analysiswasin termsof one variable, the meana. Forthe unbounded case with the restriction that σ 2 ≤ac our analysis involved two variables, a and p. Now the more general case involves four variables,a, p, c, and σ 2 . With four degrees of freedom in our equations it is not convenient to write a theorem involving these variables. A numerical investigation shows that the new product-Chebyshev bound can be better than the product bound even with σ 2 >ac. Furthermore,inactualcasesitmayturnoutthatσ 2 isnotonlylessthanacbutitmaybenoticeably less. In this case the bound will be noticeably better. Foranexample,ifwetakea=2andc =1,wefind vianumericalanalysisthatweneedp≥0.85 andatthesametimeσ 2 >actofind acasewhere thenew bound isnotbetter thanproductbound. However, when we are able to utilize σ 2 ≤ac we can derive more practical bounds to aid analysis as shown below. Above for Δ =2 we looked at the case where a6=pσ. If a =pσ we get P 2;a,p,c,σ (u)=(cp 2 σ 2 +2a 3 −2a 2 c)u−a 2 σ 2 −a 4 +2a 3 c−a 2 c 2 . Again,inthiscase wecan obtainsomebenefit withthenewbound overawiderangeofparameters. 2.4 Lower Tail Bounds 2.4.1 Bounded Case Theorem 2.4.1. Suppose X ≥ 0 with EX = a > 0 and let Y be a size biased version of X such that Y ≤X +c for some c∈ (0,∞). Let k =b |x−a| c c, I a,c = c+2a− √ c 2 +4ac 2 , a≥0. Define for 0≤x≤a, L P (x,a,c):= Y 00 and let Y be a c−bounded size biased version of X with probability p such that P(Y ≤ X +c|Y ≥ x)≥ p for some c∈ (0,∞). Further, assume that σ 2 ≤ac. Let k p be defined by k p =b |x−ap| c c and let I a,c =u 1 (as defined below). Define for 0≤x≤ap, L P,p (x,a,c):= Y 0 0.92 (for p to 2 decimal places) and has exactly one real root, otherwise. For any value of p numerical analysis shows there is exactly one real root less than ap (and positive) and is given by u 1 =A+ B 3 q 3(C+D+ p E+(F +G) 2 ) −H 3 q K +L+ p M +(N +P) 2 55 where, A = 2+2a+ap 3 B = 3 √ 2(−1+a−a 2 −4ap+2a 2 p−a 2 p 2 ) C = 2+24a+24a 2 +2a 3 −15ap−12a 2 p D = −6a 3 p−12a 2 p 2 +6a 3 p 2 −2a 3 p 3 E = 4(−1+a−a 2 −4ap+2a 2 p−a 2 p 2 ) 3 F = 2+24a+24a 2 +2a 3 −15ap−12a 2 p G = −6a 3 p−12a 2 p 2 +6a 3 p 2 −2a 3 p 3 H = 1 3 3 √ 2 K = 2+24a+24a 2 +2a 3 −15ap−12a 2 p L = −6a 3 p−12a 2 p 2 +6a 3 p 2 −2a 3 p 3 M = 4(−1+a−a 2 −4ap+2a 2 p−a 2 p 2 ) 3 K = 2+24a+24a 2 +2a 3 −15ap−12a 2 p P = −6a 3 p−12a 2 p 2 +6a 3 p 2 −2a 3 p 3 NumericalanalysisrevealsthatP 1;a,p (u)<0foru∈ (I a,c , ap)andinsertingcbackintotheanalysis establishes this part of the theorem. Now we remove the restrictionσ 2 ≤ac. In this case we have L PC,1,p (x,a,c,σ):= σ 2 σ 2 +(a−x−(k p −1)c) 2 Y 0t)≤e −t 2 /(2v) and P(Z−EZ <−t)≤e −t 2 /(2v) . We note this inequality is not in general as strong as those presented in this study. For example, if c i = 1 for all i (like in the lightbulb problem analyzed in [12]), the decay rate is less than that given originally in [12]. So, in addition to providing a framework where hard problems involving dependency can be solved (see [11] and [12] and examples below), the new inequalities here can even apply where other inequalites are also used. Example 2.6.1: Coverage Process. See Ghosh and Goldstein [11] for details of the example. Here V is the total volume of the union of n balls with radius ρ uniformly distributed in a cube C n =[0,n 1/d ) d with periodic boundary conditions whose totalvolume isn. We will letn =20,d= 2,ρ=0.5 so that φ=π/4 and μ V =n(1−(1−φ/n) n )=11.02. The maximum possible value of V in this case is V max = nφ = 15.71. Using our earlier notation c=φ, a =μ V and X =V. See below for the performance chart where we bound P(X ≥x). 59 U V U W X Y U V Z W X [ W X \ W X ] W X ^ W X _ W X ‘ W X a Y X W b Z V U b Z V c b d V U b d V c b e V U b e V c b c V U b c V c b f V U g h i j k l j m n o p i l k q l r o s t u o v l w i j n x i h j l p i t t y g n m z { i | } ~ ~ | } ~ ~ Figure 2.3: Coverage Example. Example 2.6.2: Sliding Window. See Ghosh and Goldstein [11] for details. Here for n≥m≥ 1, X is the sum of the product of m i.i.d. Bernoulli random variables in a sliding window fashion. In this example we take the simple case of m = 1 and n = 20 with p = 1/3 so the sum is actually binomial. We get c = 1 and a = 6.67. See below for the performance chart where we bound P(X ≥x). Example2.6.3: Urn allocation. See Ghosh and Goldstein[11]fordetails. Heren ballsare thrown independently into one of m urns where, for i=1,2,...,m, the probability a ball lands in the ith 60 ¡ ¢ £ ⁄ ¥ ƒ § ¤ ƒ ⁄ ƒ ' “ ƒ ⁄ « ‹ › ¤ fi fl – † ‡ · † ¶ – fi fl – † ‡ · • ‚ „ ” » … ‰ „ ” † ¶ – Figure 2.4: Sliding Window Example. urn is p i . The random variable X is the number of nonempty urns. We will consider the uniform case with p i =1/m. In this case c=2 and a =n 1− 1− 1 m n−1 ! . For analysis we will take n = 20 and m = 20. See below for the performance chart where we bound P(X ≥x). Example 2.6.4: Sliding m Window Statistics. Let n≥m≥1,V =1,...,n considered modulo n. Let V α =α,α+1,...,α+m−1, α∈V. We have max β |β :V β ∩V α 6=0|=2m−1 61 ¿ ¿ ` ´ ˆ ¿ ˜ ` ´ ¯ ` ´ ˘ ` ´ ˙ ` ´ ¨ ` ´ ` ´ ˚ ` ´ ¸ ˆ ´ ` ˜ ˝ ˛ ˝ ˇ ˝ ˝ ˝ — ˝ ˝ Æ ª ª ª Ł Ø Œ º Ø º Œ Ł Ø Œ º æ ı Ø º Œ Figure 2.5: Urn Allocation Example. and as shown in [11] we may take c=2m−1 as the bounding difference constant. Let Y be the length m runs in n trials using ζ 1 ,ζ 2 ,...,ζ n i.i.d. (0,1) Bernoulli(p) random variables. Next let X i = ζ i ,ζ i+1 ,···,ζ i+m−1 and let Y = P n i=1 X i = P n i=1 ζ i ,ζ i+1 ,···,ζ i+m−1 and then set ζ 0 j = ζ j , j6∈i,...,i+m−1 1, j∈i,...,i+m−1. Then Y s = P n i=1 ζ 0 i ,ζ 0 i+1 ,···,ζ 0 i+m−1 is the number of m runs of ζ 0 j n i=1 and Y s ≥Y. Clearly, μ=np m and as shown in [11] we have σ 2 =np m 1+2 p−p m 1−p −(2m−1)p m . 62 ł ø œ ß ł ø œ ß ł ø œ ß ł ø œ ß ł ł ø œ ł ł ! " " # $ % & ’ " ( ) ) ( * + $ ’ % , - # $ % & ’ $ & . / 0 1 ’ . ’ 2 3 . # 4 0 3 . 5 0 1 ) % 6 ( - . ’ 7 8 9 ! : 8 ; < = > ? 8 ; ? ; ; ; ? 8 @ m 8 A B < ; C = > ? s D 8 C A < C Figure 2.6: Length m Runs Example. See figure above for the performance chart where we bound P(Y ≥y). Example 2.6.5: M of N Signal Detection. Here we look at M of N logic to decide the presence or absence of a signal in noise. We let n ≥ N ≥ M ≥ 1. Let Y be the number of false alarms in n trials using ζ 1 ,ζ 2 ,...,ζ n i.i.d. (0,1) Bernoulli(p) random variables. In this context, p istheprobabilitythatthedetectedenergyexceeds somethresholdthatissettocontrolfalsealarms. Next let S i = P N−1 k=0 ζ i+k and X i = S i 1(S i ≥ M) where 1 denotes the indicator function. Then let Y = P n i=1 X i = P n i=1 1((ζ i +ζ i+1 +···+ζ i+N−1 )≥M) and ζ 0 j = ζ j , j6∈i,...,i+N−1 x j , j∈i,...,i+N−1 63 where the choice for x j is explained below. Now let Y S = P n i=1 1((ζ 0 i +ζ 0 i+1 +···+ζ 0 i+N−1 )≥M). As with the sliding m window statistics above we let c =2N−1. We see that p S =P(S i ≥M)= N X k=M N k p k (1−p) N−k . Then, clearly EY =np S and Var(Y)=np S (1−p S ). Now to set x j we proceed as follows: Let W k ={s 1 ,s 2 ,...,s T } where T = N k and each s i is a set consisting of k 1’s and (N−k) 0’s, for k = 0,1,...,N, and each s i is different. Next choose W t ,t∈{M,M +1,...,,N}at random with probability N t p t (1−p) N−t N M p M (1−p) N−M + N M+1 p M+1 (1−p) N−M−1 +···+ N N p N (1−p) N−N . Here we are conditioning on the event that we have M or more detects. After choosing W t , pick a s i ∈ W t uniformly and then let x j be the sequence of 0’s and 1’s in s i . See figure below for the performance chart where we bound P(Y ≥y) and compare various bounds. 64 E F G H I J E F G H I K L M N O P Q E F G H I R E F G H I E E F G S I I I R I K I T I U I E I I E R I E K I E T I V W X Y Z [ \ ] ^ _ ‘ a b c d e f c g h i g c j k l l l m n n o p o q r o s t u v v u w p x l y r z k x y q l w p x l y r z k x y q l s m v y { u z o l | } ~ Figure 2.7: M of N Example. 65 Bibliography 1. Abramowitz, M., and Stegun, I.: Handbook of Mathematical Functions, with Formulas, Graphs, and Mathematical Tables. Dover Publications, New York (1966) 2. Arratia R. and Baxendale P.: Bounded size bias coupling: a Gamma function bound, and universal Dickman-function behavior, Probab. Theory Related Fields, 162(3-4): 411-429, 2015. 3. Arratia R. and Goldstein, L.: Size bias, sampling, the waiting time paradox, and infinite divisibility: when is the increment independent, unpublished manuscript, 2009. 4. Arraita, R., Goldstein, L., and Langholz, B. (2005). Local central limit theorems, the high- ordercorrelationsofrejectivesamplingand logisticlikelihoodasymptotics. Ann. Statist.,33, pp. 871–914. 5. Boistard,H.,Lopuha¨ a, H.P., andRuiz-Gazen, A.: Approximationofrejectivesamplinginclu- sion probabilities and application to high order correlations. Electron. J. Stat. 6, 1967-1983 (2012). 6. BoucheronS.,Lugosi,G.andMassart,P.:ConcentrationInequalities,OxfordUniversityPress, 2013. 7. Chen, L., Goldstein, L. and Shao, Q.: Normal Approximation by Stein’s Method, Springer, 2011. 8. CookN.,GoldsteinL. andJohnson T.: Size biasedcouplingsand thespectralgap forrandom regular graphs, arXiv:1510.06013v2,Jan. 8, 2016. 66 9. Deuschel, J.-D.and Stroock,D.:Large Deviations,AMS Chelsea Publishing, American Math- ematical Society, 1989. 10. Garrappa,R.(2007)SomeFormulasforSumsofBinomialCoefficientsandGammaFunctions. International Mathematical Forum, 2, pp. 725-733. 11. Ghosh S. and Goldstein L.: Applications of size biased couplings for concentration of mea- sures, Electronic Communications in Probability, 2011. 12. GhoshS. and GoldsteinL.: Concentrationsofmeasures viasize-biased couplings, Probability Theory Related Fields, 2011. 13. Graham,R., Knuth, D.,and Patashnik,O. (1994). Concrete Mathematics: A Foundation for Computer Science. Second edition. Addison-Wesley Publishing Company, Reading, MA. 14. H´ ajek, J. (1964) Asymptotic theory of rejective sampling with varying probabilities from a finite population. Ann. Math. Statist.,35, pp. 1491–1523. 15. H´ ajek,J. (1981)Samplingfroma finitepopulation. Editedby VclavDupaˇ c. Withaforeword by P. K. Sen. Statistics: Textbooks and Monographs,37, Marcel Dekker, Inc., New York. 16. Hollander, F. den:Large Deviations, Fields Institute Monographs, American Mathematical Society, 2000. 17. Ledoux, M.: The Concentration of Measure Phenomenon, American Mathematical Society, 2001. 18. Raginsky, M. and Sason, I.: Concentration of Measure Inequalities in Information Theory, Communications and Coding, Foundations and Trends in Communications and Information Theory, vol. 10, no. 1-2, pp. 1-246, 2013. 67
Abstract (if available)
Abstract
For N=1,2,..., let S_N be a simple random sample of size n=n_N from a population A_N of size N, where 0≤ n ≤N. Then with f_N=n/N, the sampling fraction, and 1_A the inclusion indicator that A∊S_N, for any H⊂A_N of size k≥ 0, the high order correlations
Corr(k) = E ( prod_{A∊H} (1_A - f_N) )
depend only on k, and if the sampling fraction f_N → f as N → ∞, we show in this thesis that
N^(k/2) Corr(k) → [ f (f-1) ]^(k/2) EZ^k, k even
and
N^((k+1)/2) Corr(k)-> [ f(f-1) ]^((k-1)/2)(2f-1)(1/3)(k-1) EZ^(k+1), k odd
where Z is a standard normal random variable and the notation
prod_{A∊H} means the product is taken over all A in H.
We also demonstrate in this manuscript how the Chebyshev inequality can be combined with powerful product bounds to provide slightly stronger bounds than the product bounds alone under certain conditions.
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Walker, Christopher Wayne
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High order correlations in sampling and concentration bounds via size biasing
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College of Letters, Arts and Sciences
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Doctor of Philosophy
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Applied Mathematics
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2022-08
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07/25/2022
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